STRUCTURE OF ZIRCONIUM ISOTOPES
.By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( September 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Naturally occurring zirconium (Zr) is composed of four stable isotopes (of which one may in the future be found radioactive), and one very long-lived radioisotope (Zr-96), a primordial nuclide that decays via double beta decay with an observed half-life of 2.0×1019 years; it can also undergo single beta decay which is not yet observed, but the theoretically predicted value of t1/2 is 2.4×1020 years.The second most stable radioisotope is Zr-93 which has a half-life of 1.53 million years. Twenty-seven other radioisotopes have been observed. All have half-lives less than a day except for Zr-95 (64.02 days), Zr-88 (63.4 days), and Zr-89 (78.41 hours). The primary decay mode is electron capture for isotopes lighter than Zr-92, and the primary mode for heavier isotopes is beta decay. STRUCTURE OF Zr-80, Zr-82, Zr-84, Zr-86, Zr-88, Zr-90, Zr-92, Zr-94, AND Zr-96 WITH S = 0 ''' For understanding the structure of the above nuclides you must study the diagram of Zr-80 with S = 0 having 40 protons and 40 neutrons in my STRUCTURE OF Zr-90...Zr-94 . Particularly the structure of the above nuclides with extra neutrons of opposite spins is based on the structure of Zr-80 with S=0 In this structure the protons make blank positions for receiving extra neutrons which make two bonds per neutron but the small number of neutrons in the unstable Zr-82, Zr-84, Zr-86, and Zr-88 cannot give enough binding energies to pn bonds for overcoming the pp and nn repulsions. However in the stable Zr-90, Zr-92, and Zr-94 the greater number of neutrons gives enough energies to pn bonds which lead to the stability. Whereas the two more neutrons in the unstable Zr-96 under the absence of blank positions make single bonds leading to the decay. '''STRUCTURE OF Zr-98, Zr-100, Zr-102, Zr-104, Zr-106, AND Zr-108 WITH S =0 In the presence of more extra neutrons of opposite spins than those of Zr -96 the structure of the above unstable nuclides is based on the Zr-96 because the more extra neutrons than those of Zr-96 make single bonds leading to the decay. STRUCTURE OF Zr-81, Zr-83, AND Zr-103 Using again the diagram of Zr-80 we see that when the p40n40 with S=+1 goes to p37n37 with S =+1 for making horizontal bonds the change of spins gives S= -2. Then adding the one extra n41(+1/2) we get the structure of Zr-81 with S =-3/2. That is S = -2 + 1(+1/2) = -3/2 Therefore in the presence of extra neutrons we see that the structure of the above nuclides is based on the structure of Zr-81 with S =-3/2. For example the Zr-103 with S = -5/2 has two extra neutrons of negative spins and 20 extra neutrons of opposite spins giving S =0. That is the total spin of Zr-103 is given by S = -3/2 + 2(-1/2) = -5/2 ' ' STRUCTURE OF Zr-85, Zr-87, Zr-89, Zr-91, Zr-93, AND Zr-95 ' For understanding the structure of these nuclides you must read my STRUCTURE OF Zr-91 with S =+5/2. In this case we observe an opposite procedure, because the p37n37 with S = -1 goes with S = +1 to p38n38 to make horizontal bonds with the p38n38. Then adding the one extra n41(-1/2) we get a new structure of Zr-81 with S =+5/2. Under this condition in the presence of extra neutrons we see that the structure of the above nuclides is based on the new structure of Zr-81 with S = +5/2. In this structure the protons make blank positions for receiving extra neutrons which make two bonds per neutron. However in the unstable Zr-85, Zr-87, and Zr-89 the small number of neutrons cannot give enough binding energies to pn bonds for overcoming the repulsions and only in the stable structure of Zr-91 the greater number of extra neutrons is able to give enough binding energies for overcoming the repulsions. On the other hand in the unstable nuclides like the Zr-93 and Zr-95 the more extra neutrons than those of Zr-91 under the absence of blank positions make single bonds leading to the decay. ' ''' '''NUCLEAR STRUCTURE OF Zr-79 AND Zr-78 In the absence of two neutrons of opposite spins we see that the structure of the Zr-79 with S = +5/2 is based on the new structure of Zr-81 with S = +5/2. Similarly in the absence of two neutrons of opposite spins we get the structure of Zr-78 with S = 0 based on the structure of Zr-80 with S=0. ' ' NUCLEAR STRUCTURE OF Zr-97, Zr-99, AND Zr-101 After a careful analysis of the following diagram of Zr-80 we see that in the presence of a large number of extra neutrons the structure of the Zr-81 has a spin S =+1/2 because the one extra n41(+1/2) fills the blank position formed by p38 and p31. Under this condition the structure of the above nuclides is based on the new structure of the Zr-81 with S = +1/2. For example the Zr-97 and Zr-99 with S =+1/2 have 16 or 18 extra neutrons of opposite spins. Whereas the Zr-101 with S = +3/2 has 2 extra neutrons of positive spins and 18 extra neutrons of opposite spins giving S =0. That is S = +1/2 + 2(+1/2) + 0 = +3/2. ' ' DIAGRAM OF Zr-80 WITH S =0 This structure of high symmetry is based on the parallelepiped of Mg-24 having six horizontal planes of opposite spins like the +HP1, -HP2, +HP3, -HP4, +HP5, and -HP6. We also see the horizontal squares of negative spins and positive ones like the -HSQ and the +HSQ . Here the deuterons existing from p13n13 to p20n20 and the deuterons existing from p33n33 to p36n36 are not shown because they are in front of the Mg-24 or behind it. Note that all the 40 deuterons existing from p1n1 to n40p40 give a total spin S= 0. ' ' ' n40.......p40' ' +HSQ p38..........n38 ' ' n31……p12........n12......p32' ' -HP6 p31......n11.........p11…… n32 ' ' p29....... n10.........p10……n30' ' +HP5 n29……p9..........n9 …….p30 ' ' n27.........p8..........n8.......p28' ' -HP4 p27.......n7........p7.......n28 ' ' p25.........n6.........p6.......n26' ' +HP3 n25……p5........n5……...p26 ' ' n23………p4........n4……..p24' ' -HP2 p23…….n3…….p3……….n24 ' ' p21.........n2………p2........n22' ' +HP1 n21......p1........n1.........p22 ' ' p37......n37 ' ' -HSQ n39......p39 ' ' ' ' ' Category:Fundamental physics concepts